CURVED SURFACE AREA EXAMPLES

we are going to have some practice problems on curved surface-area for some particular solids.

Example 1 :

A circus tent is in the form of a cylinder with a height of 3 m and conical above it. If the base radius is 52.5 m and the slant height of the cone is 53 m, find the canvas needed to make the tent.

Solution :

Cylindrical part:

The radius of the cylindrical part (r) = 52.5 m

Height of the cylindrical part(h) = 3 m

Curved surface-area of the cylindrical part = 2πrh

= 2π(52.5)(3)

= 315 π m²

Conical part:

The radius of the conical part = 52.5 m

Slant height of the conical part = 53 m

Curved Surface-area of the conical part = πrl

= π (52.5) (53)

= (2782.5) π

Area of the canvas required :

Area of the canvas required = CSA of the cylindrical part + CSA of the conical part

= 315π + (2782.5)π

= (3097.5)π m²

Hence, area of the canvas required = (3097.5)π m²

Example 2 :

A vessel is in the form of hollow cylinder which has been surmounted on a hemispherical bowl.The diameter of a hemisphere is 14cm and the total height of a vessel is 13cm. Find the required curved surface area of the vessel.

Solution :

Diameter of the hemisphere = 14 cm

Radius of the hemisphere = 14/2 = 7 cm

Radius of the cylinder = radius of the hemisphere = 7 cm

Total height of the vessel = 14 cm

Total height of the vessel = height of the cylinder + radius of the hemisphere

13 = height of the cylinder + 7

Height of the cylinder = 13 - 7 = 6 cm

Curved Surface Area of the vessel = CSA of the cylinder + CSA of the hemisphere